In order to drive state-of-the-art high performance monitors it is necessary to use a high power and high bandwidth driver circuit capable of operating at speeds in excess of 33 MHz, while driving 1.4 Volt signals onto a 75 ohm line. Conventional driver circuits use a push-pull output stage, as illustrated in FIG. 1, with high speed npn and pnp transistors, in order to prevent large amounts of wasted power in the output stage. Within the driver circuit of FIG. 1, a collector of an npn transistor 10 is coupled to a positive supply voltage +VCC. A collector of a pnp transistor 12 is coupled to a negative supply voltage -VCC. A base of the transistor 10 is coupled to a base of the transistor 12, thereby forming an input node to which an input voltage signal Vin is coupled. An emitter of the transistor 10 is coupled to an emitter of the transistor 12 and to a first terminal of a load resistor RL, thereby forming an output node from which an output voltage signal Vout is provided. A second terminal of the resistor RL is coupled to ground.
In this driver circuit, the transistors 10 and 12 are coupled together in such a way that they cannot simultaneously conduct. When the input voltage Vin is at a level of zero volts, both of the transistors 10 and 12 are cut off and the output voltage Vout is also at a level of zero volts. As the input voltage Vin goes positive and exceeds a level of 0.5 volts, the transistor 10 begins to conduct and operates as an emitter follower. In this case, the output voltage Vout will follow the input voltage Vin and will be equal to the value of the input voltage Vin less the voltage drop across the base-to-emitter junction of the transistor 10, as illustrated by the following equation: EQU Vout=Vin-V.sub.BE10 When Vin&gt;0.5 Volts (1)
Also, in this situation, when the input voltage is greater than 0.5 volts, the transistor 10 will supply the load current IL through the load resistor RL. Because the level of the output voltage Vout is less than the level of the input voltage Vin, the emitter-base junction of the transistor 12 is reverse-biased and will be cut off. As the input voltage Vin remains positive the output voltage Vout will follow the input voltage Vin and remain at a level equal to the input voltage Vin less the voltage across the base-to-emitter junction of the transistor 10.
When the input voltage Vin goes negative and is less than -0.5 volts, the transistor 12 begins to conduct and operates as an emitter follower. Again, the output voltage Vout will follow the input voltage Vin and will be equal to the value of the input voltage Vin plus the voltage drop across the base-to-emitter junction of the transistor 12, as illustrated by the following equation: EQU Vout=Vin+V.sub.BE12 When Vin&lt;-0.5 Volts (2)
When the input voltage is less than -0.5 volts, the transistor 12 will sink the load current IL, from the load resistor RL. In this situation, the load current IL is flowing in a direction opposite to the direction shown in FIG. 1. Because the level of the output voltage Vout is greater than the level of the input voltage Vin, the emitter-base junction of the transistor 10 is reverse-biased and will be cut off. As the input voltage Vin remains negative, the output voltage Vout will follow the input voltage Vin and remain at a level equal to the input voltage Vin plus the voltage drop across the base-to-emitter junction of the transistor 12.
The transistors 10 and 12 are biased at zero current and conduct only when the input voltage signal Vin is present. The driver circuit of FIG. 1 operates in a push-pull fashion: the transistor 10 pushes or sources current into the load resistor RL when the input voltage Vin is positive and the transistor 12 pulls or sinks current from the load resistor RL when the input voltage Vin is negative. The output voltage Vout follows the input voltage Vin as it goes positive or negative.
The driver circuit of FIG. 1 with high speed npn and pnp transistors will provide a high power and high bandwidth driver circuit for driving high performance monitors. However, such a driver circuit, requiring high speed npn and pnp transistors, requires a relatively expensive process to manufacture. As is well known among those skilled in the art, such high-speed pnp transistors are expensive to include within a system or integrated circuit. What is needed is a driver circuit having low power consumption and high frequency output with high current drive which does not require such high speed pnp transistors and can therefore be manufactured from an inexpensive process.